Add the following rational expressions. Keep the denominator in your answer in its factored form. $\dfrac{4}{(8-k)(k+5)}+\dfrac{3k^2}{(6k+1)(k+5)}=$
Answer: We can add two rational expressions whose denominators are equal by adding the numerators and keeping the denominator the same. [Does this fit with how we add rational numbers?] When the denominators are not the same, we must manipulate them so that they become the same. In other words, we must find a common denominator. Since the two denominators share some of their factors, the common denominator is not simply the product of these two denominators. Instead, it's the least common multiple of the denominators: [What's that?] $(8-k)(6k+1)(k+5)$ [How did we find this?] Let's manipulate the expressions to have that denominator: $\begin{aligned}&\phantom{=}\dfrac{4}{({8-k})({k+5})}+\dfrac{3k^2}{({6k+1})({k+5})}\\\\\\\\ &=\dfrac{4({6k+1})}{({8-k})({k+5})({6k+1})}+\dfrac{3k^2({8-k})}{({6k+1})({k+5})({8-k})}\end{aligned}$ [Why did we do that?] Now that both denominators are the same, let's add! $\phantom{=}\dfrac{4(6k+1)}{(8-k)(x+5)(6k+1)}+\dfrac{3k^2(8-k)}{(6k+1)(k+5)(8-k)}$ $\begin{aligned} &=\dfrac{4(6k+1)+3k^2(8-k)}{(8-k)(k+5)(6k+1)}&\text{Add numerators}\\\\\\\\ &=\dfrac{24k+4+24k^2-3k^3}{(8-k)(k+5)(6k+1)}&\text{Expand parentheses}\\\\\\\\ &=\dfrac{-3k^3+24k^2+24k+4}{(8-k)(k+5)(6k+1)}\end{aligned}$ In conclusion, $\dfrac{4}{(8-k)(k+5)}+\dfrac{3k^2}{(6k+1)(k+5)}=\dfrac{-3k^3+24k^2+24k+4}{(8-k)(k+5)(6k+1)}$